You notice that your sample begins to boil at 96.20C and the liquid begins to drip into
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Answer:
In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.
Explanation:
The balanced reaction is
H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.
The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:
in units of
then the number of moles can be calculated as:
number of moles= molarity* volume
You have acid H₂SO₄
- 35.00 mL= 0.035 L (being 1,000 mL= 1 L)
- Molarity= 0.737 M
Then:
number of moles= 0.737 M* 0.035 L
number of moles= 0.0258
So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?
moles of KOH= 0.0516
Then 0.0516 moles of KOH are needed. So you know:
- Molarity= 0.827 M
- number of moles= 0.0516
- volume=?
Replacing in the definition of molarity:
Solving:
volume=0.0624 L= 62.4 mL
<u><em>In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.</em></u>